Question

$\forall y\in R$, f(y) =$\left|\begin{array}{ccc}1& y& y+1\\ 2y& y(y-1)& y(y+1)\\ 3y(y-1)& y(y-1)(y-2)& y(y^2-1)\end{array}\right|$ ,then
${\int }_{-\pi /2}^{\pi /2}f(y^2+2)dy$ equals

• A. $\pi$
• B. $-\pi$
• C. $0$
• D. $2-\pi$

Answer 1

Answer: (C)

$0$

$f\left(y\right)=\left|\begin{array}{ccc}1& y& y+1\\ 2y& y(y-1)& y(y+1)\\ 3y(y-1)& y(y-1)(y-2)& y(y^2-1)\end{array}\right|$
$=y(y-1)(y+1)\left|\begin{array}{ccc}1& 1& 1\\ 2y& (y-1)& y\\ 3y& (y-2)& y\end{array}\right|$ [taking common y, (y-1) and y+1 from second column, third row and third column respectively
$C_1\to C_1-C_2,C_2\to C_2-C_3$
$=y(y-1)(y+1)\left|\begin{array}{ccc}0& 0& 1\\ y+1& -1& y\\ 2y+2& -2& y\end{array}\right|$
$=y(y-1)(y+1)\left[1\right(-2y-2+2y+2\left)\right]=0$
$\Rightarrow f\left(y\right)=0$ for all $y\in R$
${\int }_{-\pi /2}^{\pi /2}f(y^2+2)dy=0$